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TODO: the source glass-rod diagram is unavailable. The recovered prose gives a surface curvature and object distance, but the diagram's orientation, labels and any further optical surfaces cannot be verified. The following solves the surviving request for a single convex entrance surface; it is conditional on that interpretation and does not reconstruct the missing diagram or supply a rod length.

For refraction at a spherical surface, take incident light travelling rightwards from refractive index $n_0$ to $n_g$, object distance $u>0$ to the left of the vertex, image distance $v$ positive to the right, and signed curvature $R_s$ positive when the centre is inside the rod. In the paraxial approximation, applying Snell's law to a light ray at height $y$ givesAfter cancelling $y$, the image position and transverse magnification areThe magnification follows by considering a paraxial light ray through the vertex: Snell's law there gives $n_0(-y_o/u)=n_g(y_i/v)$.

For a convex entrance, $R_s=+20\mathrm{mm}$ and $u=60\mathrm{mm}$. In air, the denominator $(n_g-n_0)/R_s-n_0/u$ is $1/75\mathrm{m}{\mathrm{m}}^{-1}$, givingThis is a real, inverted image within a sufficiently long glass rod, before any subsequent surface.

For water with refractive index $1.33$, the denominator is $-13/1500\mathrm{m}{\mathrm{m}}^{-1}$, andThe emerging light rays diverge as though from an upright virtual image on the object side. The change of sign is consistent with $v$ diverging when $n_0=n_{g}u/(u+R_s)=1.20$. A reversed curvature or a second exit surface changes these answers; the unavailable diagram prevents a definitive answer for the full illustrated arrangement.

Use the original diagram's positive distances: $u$ from the object to the vertex, $d$ from the vertex to the image, and $R$ from the vertex toward the centre of curvature on the outgoing side. Let a paraxial light ray strike at height $Y$. Its incoming inclination is $\alpha \simeq Y/u$ above the optical axis, its outgoing inclination is $\beta \simeq Y/d$ below it, and the inward surface normal is inclined by $\theta \simeq Y/R$ below it. Thus the incidence and refraction angles relative to that normal are respectively $\alpha +\theta$ and $\theta -\beta$.

The local tangent-plane form of Snell's law, using the small-angle approximation, givesSubstitute the three height-to-distance ratios and cancel $Y$:This refraction at a spherical surface result keeps only first-order ray angles. The surface sag shifts the height's axial reference only at higher order. Distances for other object, image or curvature orientations must be signed consistently; the displayed positive-distance form follows the particular orientation in the paper.

Keep the paper's $R_1,R_2>0$ as magnitudes for the two convex faces. With the standard signed-curvature convention measured to the right, the first surface has signed radius $+R_1$ and the second $-R_2$. This explains why the requested formula contains a sum of the two reciprocal magnitudes.

Let the axial lens thickness be $\ell$, and let $d_1$ be the distance from the first vertex to the intermediate focus that its refracted rays would reach without the second surface. The first refraction at a spherical surface givesAt the second face those rays are still converging toward that intermediate focus, so its object distance is negative, $u_2=-d_2$, with $d_2=d_1-\ell$. Applying the signed spherical-surface equation again givesIn the thin lens limit, $d_2\simeq d_1$. Adding cancels the intermediate-focus terms:The focal length is the image distance for parallel incident light rays, $u\to \infty$, hence the lensmaker's equation isThe same surrounding refractive index $n_1$ is assumed on both sides. For $n_2>n_1$ this biconvex thin lens has positive focal length. With signed radii $R_a=R_1$, $R_b=-R_2$, the equivalent familiar expression is $1/f=(n_2/n_1-1)(1/R_a-1/R_b)$.